Question

A 61.0mL sample of a 0.112M potassium sulfate solution is mixed with 35.0mL of a 0.104M lead(II) acetate solution and the following precipitation reaction occurs:

K2SO4(aq)+Pb(C2H3O2)2(aq)?2KC2H3O2(aq)+PbSO4(s)
The solid PbSO4 is collected, dried, and found to have a mass of 0.997g .
Determine the limiting reactant, the theoretical yield, and the percent yield.
Part A.
Identify the limiting reactant.
KC2H3O2
Pb(C2H3O2)2
K2SO4
PbSO4
Part B.
Determine the theoretical yield.
mass of PbSO4 =
Part C.
Determine the percent yield=

Answer 1

Lead(II) acetate is limiting reagent. (0.00364 < 0.006832)

Theoretical yield = 1.104 g

% yield = (Experimental yield / Theoretical yield) × 100 = (0.997 / 1.104) × 100 = 90.31 %

Step-by-step explanation:

Considering:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

Or,

`Moles =Molarity * {Volume\ of\ the\ solution}`

Given :

For potassium sulfate :

Molarity = 0.112 M

Volume = 61.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 61.0×10⁻³ L

Thus, moles of potassium sulfate :

`Moles=0.112 * {61.0* 10^(-3)}\ moles`

Moles of potassium sulfate = 0.006832 moles

For lead(II) acetate :

Molarity = 0.104 M

Volume = 35.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 35.0×10⁻³ L

Thus, moles of lead(II) acetate :

`Moles=0.104 * {35.0* 10^(-3)}\ moles`

Moles of lead(II) acetate = 0.00364 moles

According to the given reaction:

`KSO_4_((aq))+Pb(C_2H_3O_2)_2_((aq))\rightarrow PbSO_4_((s))+2KC_2H_3O_2_((aq))`

1 mole of potassium sulfate reacts with 1 mole of lead(II) acetate

So,

0.006832 mole of potassium sulfate reacts with 0.006832 mole of lead(II) acetate

Moles of lead(II) acetate = 0.006832 mole

Available moles of lead(II) acetate = 0.00364 mole

Limiting reagent is the one which is present in small amount. Thus, lead(II) acetate is limiting reagent. (0.00364 < 0.006832)

The formation of the product is governed by the limiting reagent. So,

1 mole of lead(II) acetate gives 1 mole of lead(II) sulfate

0.00364 mole of lead(II) acetate gives 0.00364 mole of lead(II) sulfate

Moles of lead(II) sulfate = 0.00364 moles

Molar mass of lead(II) sulfate = 303.26 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.00364 × 303.26 g = 1.104 g

Theoretical yield = 1.104 g

Given experimental yield = 0.997 g

% yield = (Experimental yield / Theoretical yield) × 100 = (0.997 / 1.104) × 100 = 90.31 %